Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are asked to estimate the smallest range of speeds that a sodium atom might have if confined to a 1.0-mm box. To do this, we need to use the Heisenberg uncertainty principle,
$$ \Delta x \cdot \Delta p_x \geq \dfrac{h}{2} $$
Hence,
$$ \Delta p_x = \dfrac{h}{2 \Delta x} $$
Plug the known;
$$ \Delta p_x = \dfrac{6.63\times 10^{-34}}{2 (1.0 \times 10^{-3})} = \bf 3.32\times 10^{-31} \; \rm{kg\cdot m/s} $$
Recalling that $ p = mv$; So the uncertainty in speed $ \Delta v_x$ is given by
$$ \Delta v_x = \frac{\Delta p_x}{m_{\rm Na}} = \frac{\Delta p_x}{ M_{\rm Na}(1.67\times 10^{-27})} $$
Plug the known;
$$ \Delta v_x = \frac{(3.32\times 10^{-31})}{(23)(1.67\times 10^{-27})} = \bf 8.64 \times 10^{-6} \; \rm{m/s} $$
Thus, the range of velocities for a sodium atom confined to a 1.0-mm box is
$$ \boxed{-4.32\times 10^{-6} \; {\rm m/s} \leq v_x \leq 4.32\times 10^{-6} \; {\rm m/s}}$$
And hence, the range of speeds for a sodium atom confined to a 1.0-mm box is
$$ \boxed{ 0\; {\rm m/s} \leq v_x \leq 4.32\times 10^{-6} \; {\rm m/s}}$$
$$\color{blue}{\bf [b]}$$
We are given that the root-mean-square (rms) speed $ v_{\rm{rms}} $ is half of the speed range we calculated in part (a).
Thus,
$$v_{\rm{rms}} = \frac{ v_x}{2} $$
Plug the known;
$$v_{\rm{rms}} = \frac{4.32\times 10^{-6}}{2} = 2.16\times 10^{-6} \; \rm{m/s} $$
Recalling that
$$ v_{\rm{rms}} = \sqrt{\frac{3k_B T}{m}} $$
So, the lowest possible temperature is given by
$$ T = \frac{m v_{\rm{rms}}^2}{3 k_B} $$
Plug the known;
$$ T = \frac{(23)(1.67\times 10^{-27})(2.16\times 10^{-6} )^2}{3(1.38 \times 10^{-23})} $$
$$T=\color{red}{\bf 4.33\times 10^{-15}}\;\rm K$$
This means that the temperature corresponding to the rms speed of the sodium atom is extremely close to absolute zero. This is well below the temperature of 1 nK mentioned in the problem. This indicates that the temperature limit set by the uncertainty principle is much lower than 1 nK.
